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......@@ -211,9 +211,9 @@
\subsection{Speed of sound}
\label{ch_speed_sound_compressibility}
An important property of fluids is the speed at which information (in particular, pressure changes due to the movement of an object) can travel within the fluid. This is equal to the average speed of molecules within the fluid, and it is called the \vocab{speed of sound}, noted~$c$.
An important property of fluids is the speed at which pressure changes can travel within the fluid (these pressure changes may for example be caused by the movement of an object). This speed is equal to the average speed of molecules within the fluid, and it is called the \vocab{speed of sound}, noted~$c$.
It is important to quantify how fast the fluid is flowing relative to e speed of sound. For this, we define the \vocab{Mach number} $\ma$ as the ratio of the local fluid speed~$V$ to the local speed of sound~$c$:
In fluid dynamics, we often quantify how fast the fluid is flowing relative to the speed of sound. For this, we define the \vocab{Mach number} $\ma$ as the ratio of the local fluid speed~$V$ to the local speed of sound~$c$:
\begin{equation}
\ma \equiv \frac{V}{c} \label{eq_def_ma}
\end{equation}
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......@@ -33,7 +33,7 @@
\label{fig_pressure_distribution_plate}
\end{figure}
\begin{enumerate}
\item What is the pressure force exerted on the left side of the plate?
\item What is the pressure force (i.e. the force resulting from the pressure) exerted on the left side of the plate?
\end{enumerate}
On the right side of the plate, the water exerts a pressure which is not uniform: it increases with depth. The relation, expressed in \si{pascals}, is:
......@@ -82,7 +82,7 @@
%homemade
\label{exo_power_lost_to_drag}
A truck moves with constant speed $\vec V$ on a road, with $V = \SI{100}{\kilo\metre\per\hour}$. Because it experiences cross-wind, it is subjected to a drag $\vec F_D$ with $F_D = \SI{5}{\kilo\newton}$ at an angle $\theta = \SI{20}{\degree}$, as shown in \cref{fig_truck_drag_power}.
A truck moves with constant speed $\vec V$ on a road, with $V = \SI{100}{\kilo\metre\per\hour}$. Because it experiences cross-wind, it is subjected to an aerodynamic force $\vec F_D$ with $F_D = \SI{5}{\kilo\newton}$ at an angle $\theta = \SI{20}{\degree}$, as shown in \cref{fig_truck_drag_power}.
\begin{figure}
\begin{center}
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......@@ -12,7 +12,7 @@
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\atstartofexercises
......@@ -89,7 +89,7 @@ Balance of energy in a considered volume with steady flow:
The pipe itself and the valve, together, induce a pressure loss which can be quantified using the dimensionless \vocab{loss coefficient} $K_\text{valve}$ (we later will later encounter it as eq.~\ref{eq_def_loss_coeff} p.\pageref{eq_def_loss_coeff}). With this tool, the pressure loss is related to the average incoming speed $V_\text{incoming}$ as:
\begin{IEEEeqnarray}{rCcCl}
K_\text{valve} &\equiv& \frac{|\Delta p_\text{valve}|}{\frac{1}{2} \rho V_\text{incoming}^2} &=& \num{2,6}
K_\text{valve} &\equiv& \frac{|\Delta p_\text{valve}|}{\frac{1}{2} \rho V_\text{incoming}^2} &=& \num{2,6}\label{eq_k_valve}
\end{IEEEeqnarray}
% Note: K = 2,6 is guesstimate: 2 for swing check valve (from White p.401) + {f L/D = 0.05 * 3 / 0.25 = 0,6}
......@@ -126,13 +126,12 @@ Balance of energy in a considered volume with steady flow:
% Fuel mass flow rough calculation: \dot Q = (\dot m * c_p * \Delta T)_air = 250 kW
% \dot Q / c_comb = 0,005 kg/s
At the outlet, the hot gases have pressure \SI{24,5}{\bar} and temperature \SI{1550}{\degreeCelsius}.
At the outlet, the hot gases have pressure \SI{24,5}{\bar} and temperature \SI{1550}{\degreeCelsius}, and exit with a speed of~\SI{50}{\metre\per\second}.
We consider that the air and gas keep the same thermodynamic properties throughout ($c_\text{p} = \SI{1050}{\joule\per\kilogram}$)
We consider that the air and gas have the same thermodynamic properties throughout the combustor ($c_\text{v} = \SI{718}{\joule\per\kilogram\per\kelvin}$, $R_\air = \SI{287}{\joule\per\kilogram\per\kelvin}$).
\begin{enumerate}
\item What is the volume flow at inlet and at outlet?
\item What is the flow velocity at outlet?
\item What is the power provided to the flow as heat?
\item What is the net force exerted on the gas as it travels through the combustor?
\end{enumerate}
......@@ -153,7 +152,7 @@ Balance of energy in a considered volume with steady flow:
\end{figure}
\begin{enumerate}
\item What is the net force exerted on the water by the wall?
\item What is the net force exerted on the water by the wall?
\item What is the net force exerted on the wall by the water?
\end{enumerate}
Now, the wall moves longitudinally in the same direction as the water jet, with a speed $V_\text{wall} = \SI{15}{\metre\per\second}$.\\
......@@ -203,7 +202,7 @@ Balance of energy in a considered volume with steady flow:
\item What is the kinetic energy per unit mass of the air at point~3?
\end{enumerate}
Once it has passed point 3, the air undergoes complex loss-inducing evolutions (including going through a \vocab{shock wave}, where its properties change very suddenly), before it discharges into the atmosphere with pressure \SI{1}{\bar} and temperature is~\SI{65}{\degreeCelsius}.
Once it has passed point 3, the air undergoes complex loss-inducing evolutions (including going through a \vocab{shock wave}, where its properties change very suddenly), before it discharges into the atmosphere with pressure \SI{1}{\bar} and temperature is~\SI{165}{\degreeCelsius}.
\begin{enumerate}
\shift{8}
\item What is the fluid velocity at outlet?
......@@ -214,13 +213,54 @@ Balance of energy in a considered volume with steady flow:
\end{enumerate}
\begin{comment}
\item [\ref{exo_water_jet}]%
\tab 1) $F_\text{net on water} = \SI{-120}{\newton}$;
\tab 2) $\vec F_\text{water/wall} = -\vec F_\text{net on water}$;
\tab 3) $F_\text{net on water} = \SI{-7,5}{\newton}$;
\end{comment}
\clearpage
\subsubsection*{Answers}
\startofanswers
\begin{enumerate}
\item p.~\pageref{exo_pipe_expansion}
\begin{enumerate}
\item $\dot m = \SI{7,53}{\kilogram\per\second}$; $\dot \vol = \SI{7,53}{\liter\per\second}$
\itme $V_2 = \SI{0,375}{\metre\per\second}$
\item $\Delta p_{1\to 2} = \SI{+1054}{\pascal}$
\item $\Delta p_{3\to 4} = \SI{+4218}{\pascal}$
\item $\Delta p_{5\to 6} = \SI{+1,29}{\pascal}$
\end{enumerate}
\item p.~\pageref{exo_pipe_with_losses}
\begin{enumerate}
\item $V_2 = V_1 = \SI{1,5}{\metre\per\second}$ by application of the mass balance equation; although a mis-application of the energy equation would suggest otherwise
\item With eq.~\ref{eq_k_valve}, $\Delta p_\text{valve} = \SI{-2925}{\pascal}$
\item $\dot W_\text{injection} = \SI{-215,37}{\watt}$
\item With eq.~\ref{eq_sfee}, $\Delta T = \SI{+7}{\milli\kelvin}$ (very small!)
\end{enumerate}
\item p.~\pageref{exo_combustor}
\begin{enumerate}
\item $\dot \vol_1 = \SI{0,0759}{\metre\cubed\per\second}$ \& $\dot \vol_2 = \SI{0,1078}{\metre\cubed\per\second}$ (there is no volume balance equation!)
\item $\dot Q = \SI{+261}{\kilo\watt}$ (using $V_2 = \SI{50}{\metre\per\second}$)
\item $F_\net = \SI{+19,25}{\newton}$ (in flow-wise direction)
\end{enumerate}
\item p.~\pageref{exo_water_jet}
\begin{enumerate}
\item $F_\text{net on water} = \SI{-120}{\newton}$
\item $\vec F_\text{water/wall} = -\vec F_\text{net on water}$
\item $F_\text{net on water} = \SI{-7,5}{\newton}$
\item $\dot W_\text{wall} = \SI{112,5}{\watt}$
\end{enumerate}
\item p.~\pageref{exo_high_speed_gas_flow}
\begin{enumerate}
\item With eq.~\ref{eq_sfee}, $V_2 = \SI{417,2}{\metre\per\second}$, and so $\dot m_2 = \dot m = \SI{14,1}{\kilogram\per\second}$
\item $\ma_1 = \num{0}$ (zero velocity) and $\ma = \num{1}$ (a classical feature of compressible flow expansions)
\item $F_{\net 1 \to 2} = \SI{+5,882}{\kilo\newton}$ (flow-wise)
\item $e_{k2} = \SI{87,03}{\kilo\joule\per\kilogram}$
\item With eq.~\ref{eq_sfee}, $V_3 = \SI{638,71}{\metre\per\second}$ (you may then calculate $\rho_3$ and obtain $A_3 > A_2$ even though $V_3 > V_2$, a classical feature of supersonic flows)
\item $\ma_3 = \num{1,795}$
\item $F_{\net 2 \to 3} = \SI{+3,137}{\kilo\newton}$
\item $e_{k3} = \SI{204,61}{\kilo\joule\per\kilogram}$
\item $V_4 = \SI{405}{\metre\per\second}$
\item $\rho_4 = \SI{0,7952}{\kilogram\per\second}$ and so $A_4 = \SI{0,0438}{\metre\squared}$
\item $\ma_4 = \num{0,96}$
\item $F_{\net 3 \to 4} = \SI{-3,309}{\kilo\newton}$ (so, against the flow direction)
\item $e_{k4} = \SI{82,01}{\kilo\joule\per\kilogram}$
\end{enumerate}
\end{enumerate}
\atendofexercises
......@@ -7,8 +7,6 @@
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\begin{IEEEeqnarray}{CCCCCCC}
\timederivative{m e_\sys} = \Sigma \left(\dot Q + \dot W\right) &=& \timederivative{} \left(m e\right)_\cv & + & \left(\dot m e\right)_\net \nonumber\\
\end{IEEEeqnarray}\pause
......@@ -352,7 +351,7 @@
\oldfigureframe{}{concept_control_volume_system_simple_2}{0.8}{\wcfile{System control volume intregral analysis.svg}{Figure} \cczero \oc}
\begin{frame}
\setcounter{equation}{15}
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What is the specific energy $e$ made of?
\begin{IEEEeqnarray}{rCl}
e &\equiv& i + e_k + e_p
......@@ -409,7 +408,7 @@
\end{frame}
\begin{frame}
Thermodynamicists don’t know everything
Thermodynamicists don’t know everything.
\begin{enumerate}
\shift{1}
\item The terms are \emph{disproportionate}.\\\pause
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......@@ -15,7 +15,7 @@
\youtubethumb{1LXlFVtPoCY}{pre-lecture briefing for this chapter, part 1/2}{\oc (\ccby)}
In this chapter, we use the same tools that we developed in \chaptertwoshort, but we are able to develop and apply them to more complex cases. Specifically, we would like to answer the following questions:
\begin{enumerate}
\item What the mass flows and forces involved when a flow is non-uniform?
\item What are the mass flows and forces involved when a flow has non-uniform velocity?
\item What are the forces and moments involved when a flow changes direction?
\end{enumerate}
......@@ -24,7 +24,7 @@
\subsection{Control volume}
Let us begin, this time, by building a control volume in \emph{any arbitrary flow}: we are no longer limited to one-inlet, one-oulet steady situtations. Instead, we will write equations that work inside any generic velocity field $\vec V = (u, v, w)$ which is a function of space and time ($\vec V = f(x, y, z, t)$).
Let us begin, this time, by building a control volume in \emph{any arbitrary flow}: we are no longer limited to one-inlet, one-oulet steady situtations. Instead, we will write equations that work inside any generic velocity field $\vec V = (u, v, w)$ which is a function of space and time :$\vec V = f(x, y, z, t)$.
Within this flow, we draw an arbitrary volume named \vocab{control volume} (CV) which is free to move and change shape (\cref{fig_cv}). We are going to measure the properties of the fluid at the borders of this volume, which we call the \vocab{control surface} (CS), in order to compute the net effect of the flow through the volume.
......@@ -124,7 +124,7 @@
\begin{IEEEeqnarray}{rCl}
0 & = & \timederivative{} \iiint_\cv \rho \diff \vol + \sum_\out \left\{ \rho V_\perp A\right\} + \sum_\inn \left\{ \rho V_\perp A\right\}\label{eq_rtt_mass_simple}\\
& = & \timederivative{} \iiint_\cv \rho \diff \vol + \sum_\out \left\{ \rho |V_\perp| A\right\} - \sum_\inn \left\{ \rho |V_\perp| A\right\} \nonumber\\
& = & \timederivative{} m_\cv + \sum_\out \left\{ \dot m \right\} - \sum_\inn \left\{ \dot m \right\}\label{eq_rtt_mass_simple_two}
& = & \timederivative{} m_\cv + \sum_\out \left\{ |\dot m| \right\} - \sum_\inn \left\{ |\dot m| \right\}\label{eq_rtt_mass_simple_two}
\end{IEEEeqnarray}
\begin{figure}[ht]
\begin{center}
......@@ -207,7 +207,7 @@
\section{Balance of angular momentum}
\youtubethumb{nmEe7Dq01AU}{pre-lecture briefing for this chapter, part 2/2}{\oc (\ccby)}
What is the moment (the “spinning effort”) applying to a fluid flowing through any arbitrary volume? We answer this question by writing an angular momentum balance in the template provided by the Reynolds transport theorem (eq. \ref{eq_rtt}).
What is the moment (the “twisting effort”) applying to a fluid flowing through any arbitrary volume? We answer this question by writing an angular momentum balance in the template provided by the Reynolds transport theorem (eq. \ref{eq_rtt}).
We now state that the placeholder variable $B$ is angular momentum $\vec r_{\X m} \wedge m \vec V$. It follows that $\inlinetimederivative{B}$ becomes $\inlinetimederivative{\vec r_{\X m} \wedge m \vec V_\sys}$, which by definition is the net moment $\vec M_\net$ applying to the system (see eq.~\ref{eq_secondlawmom} p.\pageref{eq_secondlawmom}). Also, $b \equiv B/m = \vec r_{\X m} \wedge \vec V$ and now the Reynolds transport theorem becomes:\dontbreakpage
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